p, q are two prime numbers. For n=pq,if the e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The general term $T_{r+1}$ in the expansion of $\left(x^{-5/4} + 2x^{4/5}\right)^n$ is given by:$T_{r+1} = {}^nC_r (x^{-5/4})^{n-r} (2x^{4/5})^r =

Vedclass · Accepted Answer

$82$

Vedclass · Answer

(A) The general term $T_{r+1}$ in the expansion of $\left(x^{-5/4} + 2x^{4/5}ight)^n$ is given by:$T_{r+1} = {}^nC_r (x^{-5/4})^{n-r} (2x^{4/5})^r = {}^nC_r 2^r x^{-\frac{5n}{4} + \frac{5r}{4} + \frac{4r}{5}} = {}^nC_r 2^r x^{\frac{-25n + 41r}{20}}$.For the coefficient of $x^0$ to be non-zero,the exponent must be zero:$-25n + 41r = 0 \Rightarrow r = \frac{25n}{41}$.Since $r$ is an integer and $0 \le r \le n$,$n$ must be a multiple of $41$. Given $n = pq$ where $p, q$ are primes,$n$ could be $41 	imes 1 = 41$ (not possible as $1$ is not prime) or $41 	imes 2 = 82$.For the coefficient of $x^{-n}$ to be non-zero:$\frac{-25n + 41r}{20} = -n$ $\Rightarrow -25n + 41r = -20n$ $\Rightarrow 41r = 5n$ $\Rightarrow r = \frac{5n}{41}$.For $r$ to be an integer,$n$ must be a multiple of $41$. The smallest such $n = pq$ is $41 	imes 2 = 82$.

$p, q$ are two prime numbers. For $n=pq$,if the expansion $\left(x^{-5/4} + 2x^{4/5}\right)^n$ contains non-zero coefficients of $x^{-n}$ and $x^0$,then the least value of such $n$ is

Similar Questions

The coefficient of $t^{12}$ in the expansion of $(1 + t^2)^6(1 + t^6)(1 + t^{12})$ is:

If the $2^{\text{nd}}$,$3^{\text{rd}}$,and $4^{\text{th}}$ terms in the expansion of $(x+a)^{n}$ are $96, 216$,and $216$ respectively,and $n$ is a positive integer,then $a+x=$

Suppose $A$ and $B$ are the coefficients of $30^{\text{th}}$ and $12^{\text{th}}$ terms respectively in the binomial expansion of $(1+x)^{2n-1}$. If $2A = 5B$,then $n$ is equal to:

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

$A$ person takes an examination consisting of four papers,each with a maximum of $m$ marks. The number of ways in which one can obtain a total of $2m$ marks is

Vedclass Test Series

Exam Paper Generator

Online Exam Module